A molecule in a gas container hits a horizontal wall with speed $200\ \text{m s}^{-1}$ and angle $30^\circ$ with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?

> [!TLDR] > The collision is elastic (kinetic energy is conserved) and momentum is conserved for the molecule–wall system, but not for the molecule alone — the wall’s huge mass makes its velocity change negligible, so the molecule’s momentum component normal to the wall reverses while the tangential component stays unchanged. The key to this problem is to separate the collision into two perpendicular directions: normal (perpendicular) to the wall and tangential (parallel) to the wall. The wall is essentially immovable — its mass is enormous compared to a single molecule — so it can absorb or supply momentum without any noticeable change in its own motion. Let’s work through it step by step. --- 1. **Set up the velocity components before and after the collision** The molecule hits the wall with speed $v = 200\ \text{m/s}$ at an angle $30^\circ$ to the normal. Take the normal direction as the $x$-axis (pointing into the wall) and the tangential direction as the $y$-axis (along the wall). Before collision: \[ v_{x,i} = v \cos 30^\circ = 200 \times \frac{\sqrt{3}}{2} = 100\sqrt{3}\ \text{m/s} \] \[ v_{y,i} = v \sin 30^\circ = 200 \times \frac{1}{2} = 100\ \text{m/s} \] After collision, the speed is the same ($200\ \text{m/s}$) and the angle with the normal is still $30^\circ$, but now the molecule moves *away* from the wall. So the normal component reverses sign, while the tangential component stays the same: \[ v_{x,f} = -100\sqrt{3}\ \text{m/s}, \qquad v_{y,f} = 100\ \text{m/s} \] 2. **Check kinetic energy — is the collision elastic?** Kinetic energy depends only on speed, not direction. Since the speed is $200\ \text{m/s}$ both before and after, the kinetic energy is unchanged: \[ \frac{1}{2} m v^2 \quad \text{(same before and after)} \] Therefore the collision is **elastic**. > [!WARNING] > A common mistake is to think that because the wall doesn’t move, the collision must be inelastic. But “ela

A molecule in a gas container hits a horizontal wall with speed 200m s⁻¹ and angle 30^° with the normal, and r

Exercises · 5.14

200\ \text{m s}^{-1}

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Concept understanding — Impulse Momentum Theorem

The Intuition: Why Do We Need a "New" Idea?

Imagine you're catching a cricket ball. If you let your hands stay rigid, the ball stings and might bounce off. But if you give with the ball — pulling your hands back as you catch — the catch feels soft and the ball stops gently.

Same ball, same speed, same change in momentum. But the force you feel is completely different. Why?

The answer is time. When you pull your hands back, you increase the time over which the ball slows down. A longer time means a smaller force — even though the total "oomph" needed to stop the ball is the same. That "oomph" is called impulse.

Note

Impulse is not a mysterious new quantity. It's just force multiplied by the time it acts. If you push gently for a long time, or push hard for a short time, you can produce the same effect.

The Precise Statement

The Impulse-Momentum Theorem says:

The impulse delivered to an object equals the change in its momentum.

In symbols:

\vec{J} = \Delta \vec{p}

Where:

$\vec{J}$ is the impulse (a vector)
$\Delta \vec{p}$ is the change in momentum (also a vector)

And since impulse is force times time:

\vec{F}_{\text{avg}} \, \Delta t = m \vec{v}_f - m \vec{v}_i

\vec{J} = \vec{F}_{\text{avg}} \, \Delta t = \Delta \vec{p}

Breaking It Down Piece by Piece

Momentum ( $\vec{p}$ ) is mass times velocity: $\vec{p} = m\vec{v}$ . It's a measure of how hard it is to stop a moving object. A truck moving slowly has large momentum; a bullet moving fast has large momentum too.

Impulse ( $\vec{J}$ ) is the product of the average force and the time interval over which it acts: $\vec{J} = \vec{F}_{\text{avg}} \, \Delta t$ .

The theorem connects them: the net impulse changes the momentum. If you apply a net force to an object for some time, its momentum changes by exactly that amount.

Watch out

A common mistake is to think impulse is just force. It's force × time. A huge force acting for a tiny time (like a bat hitting a ball) can produce the same impulse as a tiny force acting for a long time (like a gentle push).

Why This Matters: Real-World Examples

Catching a ball (soft vs. hard hands)

Hard hands: $\Delta t$ is small → $\vec{F}_{\text{avg}}$ is large (it hurts)
Soft hands: $\Delta t$ is large → $\vec{F}_{\text{avg}}$ is small (it's comfortable)
In both cases, $\Delta \vec{p}$ is the same (ball goes from moving to stopped)

Airbags in cars

Without airbag: your head hits the dashboard in ~0.01 s → huge force
With airbag: your head decelerates over ~0.1 s → force is 10 times smaller
Same change in momentum, but the airbag extends the time

A cricket bat hitting a ball

The bat is in contact with the ball for a few milliseconds
The force during that contact is enormous (hundreds of Newtons)
The impulse changes the ball's momentum from one direction to another

The Mathematical Derivation (Short)

Start from Newton's second law:

\vec{F}_{\text{net}} = m \vec{a} = m \frac{d\vec{v}}{dt}

Multiply both sides by $dt$ :

\vec{F}_{\text{net}} \, dt = m \, d\vec{v}

Integrate over the time interval:

\int_{t_i}^{t_f} \vec{F}_{\text{net}} \, dt = m \int_{\vec{v}_i}^{\vec{v}_f} d\vec{v} = m\vec{v}_f - m\vec{v}_i

The left side is the impulse (the area under the force-time graph). The right side is the change in momentum. …

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The collision is elastic (kinetic energy is conserved) and momentum is conserved for the molecule–wall system, but not for the molecule alone — the wall’s huge mass makes its velocity change negligible, so the molecule’s momentum component normal to the wall reverses while the tangential component stays unchanged.

The key to this problem is to separate the collision into two perpendicular directions: normal (perpendicular) to the wall and tangential (parallel) to the wall. The wall is essentially immovable — its mass is enormous compared to a single molecule — so it can absorb or supply momentum without any noticeable change in its own motion.

Let’s work through it step by step.

Set up the velocity components before and after the collision

The molecule hits the wall with speed $v = 200\ \text{m/s}$ at an angle $30^\circ$ to the normal.

Take the normal direction as the $x$ -axis (pointing into the wall) and the tangential direction as the $y$ -axis (along the wall).

Before collision:

v_{x,i} = v \cos 30^\circ = 200 \times \frac{\sqrt{3}}{2} = 100\sqrt{3}\ \text{m/s}

v_{y,i} = v \sin 30^\circ = 200 \times \frac{1}{2} = 100\ \text{m/s}

After collision, the speed is the same ( $200\ \text{m/s}$ ) and the angle with the normal is still $30^\circ$ , but now the molecule moves away from the wall. So the normal component reverses sign, while the tangential component stays the same:

v_{x,f} = -100\sqrt{3}\ \text{m/s}, \qquad v_{y,f} = 100\ \text{m/s}

Check kinetic energy — is the collision elastic?

Kinetic energy depends only on speed, not direction. Since the speed is $200\ \text{m/s}$ both before and after, the kinetic energy is unchanged:

\frac{1}{2} m v^2 \quad \text{(same before and after)}

Therefore the collision is elastic.

Watch out

A common mistake is to think that because the wall doesn’t move, the collision must be inelastic. But “elastic” only means kinetic energy is conserved — and here it clearly is. The wall’s huge mass means it can take momentum without gaining noticeable kinetic energy.

Check momentum — is it conserved? …

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Q.A molecule in a gas container hits a horizontal wall with speed $200\ \text{m s}^{-1}$ and angle $30^\circ$ with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?

Concept understanding — Impulse Momentum Theorem

The Intuition: Why Do We Need a "New" Idea?

The Precise Statement

Breaking It Down Piece by Piece

Why This Matters: Real-World Examples

The Mathematical Derivation (Short)

Concept: Elastic Collision with an Immovable Wall

Q.A molecule in a gas container hits a horizontal wall with speed 200 m s−1200\ \text{m s}^{-1}200 m s−1 and angle 30∘30^\circ30∘ with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?

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Q.A molecule in a gas container hits a horizontal wall with speed $200\ \text{m s}^{-1}$ and angle $30^\circ$ with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?

Unlock everything free for 14 days